The present invention relates to crosstalk removal technology; and more particularly, to a crosstalk removal apparatus for a semiconductor memory device, which is capable of removing crosstalk that occurs due to coupling between transmission lines.
In semiconductor memory devices, details regarding electromagnetic fields that affect a signal when it is transmitted via a transmission line will be described below.
When a dielectric is coupled between metals, there is capacitance between them. That is, since a transmission line is made of metal and a dielectric is arranged between transmission lines, capacitance exists between them. It is known that even air is a dielectric, which has a relative dielectric constant equal to ‘1’. Therefore, even when transmission lines are arranged in the air, there is capacitance between them. When an alternating current in a high frequency band flows onto a transmission line, electrical interference occurs due to the influence of capacitance between transmission lines, i.e., mutual capacitance, as the current goes toward a higher frequency band. This also affects other electrical characteristics such as the characteristic impedance value of a transmission line. In addition, as an alternating current flows onto a transmission line, a magnetic field is induced, which produces mutual inductance affecting the magnetic field of another transmission line. This mutual inductance causes magnetic energy interference that affects the inductance value of each transmission line, etc., and thus affects the characteristic impedance value of each transmission line.
Meanwhile, coupling is a phenomenon in which alternate energies such as an electric field and a magnetic field are mutually transferred between separated spaces or transmission lines. When metals are brought close to each other, this phenomenon gives rise to interference with a signal that acts as an undesired parasitic effect. In view of ElectricMagnetic Interference (EMI), this unnecessary coupling is often called “crosstalk.” In this description of the present invention, therefore, unnecessary mutual interference caused by coupling is defined as crosstalk, as will be set forth below.
Crosstalk occurs by mutual inductance and mutual capacitance. Such mutual inductance and mutual capacitance influence the total inductance and the total capacitance of a transmission line. The following is an explanation of the influence of mutual inductance and mutual capacitance in each mode, in an analysis of an ODD mode and EVEN mode in a coupled transmission line theory.
FIG. 3 shows an equivalent circuit model of a coupled transmission line, which has an equivalent circuit 310 of inductance and capacitance, an inductance equivalent circuit 320 and a capacitance equivalent circuit 330.
Signal modes between adjacent transmission lines can be largely classified into an ODD mode and an EVEN mode. When there are two transmission lines, the ODD mode and the EVEN mode are the ones when signals having a 180-degree phase difference and the same phase are applied to the two transmission lines, respectively. First, in the case of inductance, a voltage is generated by inductive coupling and currents I1 and I2 flowing onto two transmission lines in the inductance equivalent circuit 320 have the same amplitude and opposite directions to each other. Assuming that the self inductance L11=L22=L0 and the mutual inductance L12=LM, V1 and V2 of the inductance equivalent circuit 320 may be expressed as:
                              V          1                =                                            L              0                        ⁢                                          ⅆ                                  I                  1                                                            ⅆ                t                                              +                                    L              M                        ⁢                                          ⅆ                                  I                  2                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          1          )                                                  V          2                =                                            L              0                        ⁢                                          ⅆ                                  I                  2                                                            ⅆ                t                                              +                                    L              M                        ⁢                                          ⅆ                                  I                  1                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          2          )                    
In the ODD mode, I1=−I2, and V1=−V2, and therefore, these equations may be represented as follows:
                              V          1                =                                                            L                0                            ⁢                                                ⅆ                                      I                    1                                                                    ⅆ                  t                                                      +                                          L                M                            ⁢                                                ⅆ                                      (                                          -                                              I                        1                                                              )                                                                    ⅆ                  t                                                              =                                    (                                                L                  0                                ⁢                                  -                                ⁢                                  L                  M                                            )                        ⁢                                          ⅆ                                  I                  1                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          3          )                                                  V          2                =                                                            L                0                            ⁢                                                ⅆ                                      I                    2                                                                    ⅆ                  t                                                      +                                          L                M                            ⁢                                                ⅆ                                      (                                          -                                              I                        2                                                              )                                                                    ⅆ                  t                                                              =                                    (                                                L                  0                                -                                  L                  M                                            )                        ⁢                                          ⅆ                                  I                  2                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          4          )                                                  L          ODD                =                              L            11                    ⁢                      -                    ⁢                      L            M                                              Eq        .                                  ⁢                  (          5          )                    
As can be seen from Eq. (5) above, the total inductance LODD in the ODD mode becomes smaller than the self-inductance L11 by the mutual inductance LM.
Similarly, assuming that in the capacitance equivalent circuit 330 the self capacitance C1G=C2G=C0 and the mutual capacitance C12=CM, I1 and I2 of the capacitance equivalent circuit 330 may be expressed as:
                              I          1                =                                                            C                0                            ⁢                                                ⅆ                                      V                    1                                                                    ⅆ                  t                                                      +                                          C                M                            ⁢                                                ⅆ                                      (                                                                  V                        1                                            -                                              V                        2                                                              )                                                                    ⅆ                  t                                                              =                                                    (                                                      C                    0                                    ⁢                                      -                                    ⁢                                      C                    M                                                  )                            ⁢                                                ⅆ                                      V                    1                                                                    ⅆ                  t                                                      -                                          C                M                            ⁢                                                ⅆ                                      V                    2                                                                    ⅆ                  t                                                                                        Eq        .                                  ⁢                  (          6          )                                                  I          2                =                                                            C                0                            ⁢                                                ⅆ                                      V                    2                                                                    ⅆ                  t                                                      +                                          C                M                            ⁢                                                ⅆ                                      (                                                                  V                        2                                            ⁢                                              -                                            ⁢                                              V                        1                                                              )                                                                    ⅆ                  t                                                              =                                                    (                                                      C                    0                                    ⁢                                      -                                    ⁢                                      C                    M                                                  )                            ⁢                                                ⅆ                                      V                    2                                                                    ⅆ                  t                                                      -                                          C                M                            ⁢                                                ⅆ                                      V                    1                                                                    ⅆ                  t                                                                                        Eq        .                                  ⁢                  (          7          )                    
In the ODD mode, I1=−I2, and V1=−V2, and therefore, these equations may be defined as:
                              I          1                =                                                            C                0                            ⁢                                                ⅆ                                      V                    1                                                                    ⅆ                  t                                                      +                                          C                M                            ⁢                                                ⅆ                                      (                                                                  V                        1                                            -                                              (                                                  -                                                      V                            1                                                                          )                                                              )                                                                    ⅆ                  t                                                              =                                    (                                                C                  0                                +                                  2                  ⁢                                      C                    M                                                              )                        ⁢                                          ⅆ                                  V                  1                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          8          )                                                              I            2                    =                                                                      C                  0                                ⁢                                                      ⅆ                                          V                      2                                                                            ⅆ                    t                                                              +                                                C                  M                                ⁢                                                      ⅆ                                          (                                                                        V                          2                                                -                                                  (                                                      -                                                          V                              2                                                                                )                                                                    )                                                                            ⅆ                    t                                                                        =                                          (                                                      C                    0                                    +                                      2                    ⁢                                          C                      M                                                                      )                            ⁢                                                ⅆ                                      V                    2                                                                    ⅆ                  t                                                                    ⁢                                  ⁢                              C            ODD                    =                                                    C                                  1                  ⁢                  G                                            +                              2                ⁢                                  C                  M                                                      =                                          C                11                            +                              C                M                                                                        Eq        .                                  ⁢                  (          9          )                                                  C          11                =                              C                          1              ⁢              G                                +                      C            M                                              Eq        .                                  ⁢                  (          10          )                    
As can be seen from Eq. (10) above, the total capacitance CODD in the ODD mode becomes greater than the self-capacitance C1G by the mutual capacitance 2CM.
With the total inductance LODD and the total capacitance CODD in Eqs. (5) and (10) above, ZODD and TDODD may be defined as:
                              Z          ODD                =                                                            L                ODD                                            C                ODD                                              =                                                                      L                  11                                -                                  L                  12                                                                              C                  11                                +                                  C                  12                                                                                        Eq        .                                  ⁢                  (          11          )                                                  TD          ODD                =                                                            L                ODD                            ⁢                              C                ODD                                              =                                                    (                                                      L                    11                                    -                                      L                    12                                                  )                            ⁢                              (                                                      C                    11                                    +                                      C                    12                                                  )                                                                        Eq        .                                  ⁢                  (          12          )                    
The EVEN mode is one in which signals having the same phase and the same amplitude are applied to the two transmission lines, respectively. First, in the case of inductance, a voltage is generated by an inductive coupling and currents I1 and I2 flowing onto the two transmission lines in the inductance equivalent circuit 320 have the same amplitude and the same directions. Assuming that the self-inductance L11=L22=L0 and the mutual inductance L12=LM, V1 and V2 of the inductance equivalent circuit 320 may be expressed as Eqs. (1) and (2) above. And, in the ODD mode, I1=I2, and V1=V2, and therefore, these equations may be represented as:
                              V          1                =                                                            L                0                            ⁢                                                ⅆ                                      I                    1                                                                    ⅆ                  t                                                      +                                          L                M                            ⁢                                                ⅆ                                      (                                          I                      1                                        )                                                                    ⅆ                  t                                                              =                                    (                                                L                  0                                ⁢                                  +                                ⁢                                  L                  M                                            )                        ⁢                                          ⅆ                                  I                  1                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          13          )                                                  V          2                =                                                            L                0                            ⁢                                                ⅆ                                      I                    2                                                                    ⅆ                  t                                                      +                                          L                M                            ⁢                                                ⅆ                                      (                                          I                      2                                        )                                                                    ⅆ                  t                                                              =                                    (                                                L                  0                                +                                  L                  M                                            )                        ⁢                                          ⅆ                                  I                  2                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          14          )                                                  L          EVEN                =                              L            11                    ⁢                      +                    ⁢                      L            M                                              Eq        .                                  ⁢                  (          15          )                    
As can be seen from Eq. (15) above, the total inductance LEVEN in the EVEN mode has the self-inductance L11 added to the mutual inductance LM.
Similarly, assuming that in the capacitance equivalent circuit 330 the capacitance may be expressed by Eqs. (6) and (7) above, and in the EVEN mode, I1=I2, and V1=V2, and therefore, I1 and I2 may be represented again as:
                              I          1                =                                                            C                0                            ⁢                                                ⅆ                                      V                    1                                                                    ⅆ                  t                                                      +                                          C                M                            ⁢                                                ⅆ                                      (                                                                  V                        1                                            -                                              V                        1                                                              )                                                                    ⅆ                  t                                                              =                                    C              0                        ⁢                                          ⅆ                                  V                  1                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          16          )                                                              I            2                    =                                                                      C                  0                                ⁢                                                      ⅆ                                          V                      2                                                                            ⅆ                    t                                                              +                                                C                  M                                ⁢                                                      ⅆ                                          (                                                                        V                          2                                                -                                                  V                          2                                                                    )                                                                            ⅆ                    t                                                                        =                                          C                0                            ⁢                                                ⅆ                                      V                    2                                                                    ⅆ                  t                                                                    ⁢                                  ⁢                              C            EVEN                    =                                    C                              1                ⁢                G                                      =                                          C                11                            -                              C                M                                                                        Eq        .                                  ⁢                  (          17          )                                                  C          11                =                              C                          1              ⁢              G                                +                      C            M                                              Eq        .                                  ⁢                  (          18          )                    
Accordingly, the total capacitance CEVEN in the EVEN mode becomes equal to the self-capacitance C1G, as shown in Eq. (18) above. Using the total inductance LEVEN and the total capacitance CEVEN in Eqs. (15) and (18) above, ZEVEN and TDEVEN may be defined as:
                              Z          EVEN                =                                                            L                EVEN                                            C                EVEN                                              =                                                                      L                  11                                +                                  L                  12                                                                              C                  11                                -                                  C                  12                                                                                        Eq        .                                  ⁢                  (          19          )                                                  TD          EVEN                =                                                            L                EVEN                            ⁢                              C                EVEN                                              =                                                    (                                                      L                    11                                    +                                      L                    12                                                  )                            ⁢                              (                                                      C                    11                                    -                                      C                    12                                                  )                                                                        Eq        .                                  ⁢                  (          20          )                    
As mentioned above, the characteristic impedance of a transmission line varies depending on adjacent transmission lines and a signal mode due to the influence of coupling between the adjacent transmission lines, which causes a difference in the transmission rates of signals. As a result, this transmission rate difference acts as a factor that impairs a timing margin.
There are various methods to compensate the discrepancy in the transmission rate of signal, i.e., skew. As one of these methods, a Delayed Locked Loop (DLL) is frequently used for the purpose. However, devices using such a method require numerous control signals therein and thus occupy a complex and wider area, together with much power consumption for such reasons.